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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 11, Problema 11.2.51

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.


g(x) = 2/(1 − 2x)² using f(x) = 1/(1 − 2x)

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1
Recall the given function f(x) = \(\frac{1}{1 - 2x}\), which can be expressed as a geometric series centered at 0. The general form of a geometric series is \(\sum\)_{n=0}^\(\infty\) r^n = \(\frac{1}{1-r}\) for |r| < 1. Here, r = 2x, so the power series for f(x) is \(\sum\)_{n=0}^\(\infty\) (2x)^n = \(\sum\)_{n=0}^\(\infty\) 2^n x^n, valid for |2x| < 1 or |x| < \(\frac{1}{2}\).
Notice that g(x) = \(\frac{2}{(1 - 2x)^2}\) is related to f(x) by differentiation. Specifically, the derivative of f(x) with respect to x is f'(x) = \(\frac{d}{dx}\) \(\left\)( \(\frac{1}{1 - 2x}\) \(\right\)). Use the chain rule to find f'(x).
Calculate f'(x): Since f(x) = (1 - 2x)^{-1}, then f'(x) = -1 \(\cdot\) (1 - 2x)^{-2} \(\cdot\) (-2) = \(\frac{2}{(1 - 2x)^2}\). This shows that g(x) = f'(x).
Differentiate the power series for f(x) term-by-term to find the power series for g(x). The term-by-term differentiation of \(\sum\)_{n=0}^\(\infty\) 2^n x^n is \(\sum\)_{n=1}^\(\infty\) n 2^n x^{n-1}. This series represents g(x).
Determine the interval of convergence for g(x). Since differentiation does not change the radius of convergence, the interval of convergence for g(x) is the same as for f(x), which is |x| < \(\frac{1}{2}\).

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Power Series Representation

A power series expresses a function as an infinite sum of terms involving powers of the variable, typically centered at a point (here, 0). Understanding how to write functions like f(x) = 1/(1 − 2x) as a power series is essential, as it forms the basis for manipulating and finding related series for functions like g(x).
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Term-by-Term Differentiation and Integration of Power Series

Power series can be differentiated or integrated term-by-term within their interval of convergence. This property allows us to find the power series for g(x) by differentiating or integrating the known series for f(x), enabling the construction of new series representations from existing ones.
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Interval of Convergence

The interval of convergence is the set of x-values for which a power series converges. When differentiating or integrating a power series, the interval of convergence may change or remain the same, so determining this interval for the resulting series is crucial to ensure the series accurately represents the function.
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Interval of Convergence